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There is one on-line at: http://www.calculatorsoup.com/calculators/math/fractions.php 

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why not...

A car in town X travels 120KM at N60'E to town Y and then 50Km at N30'W to town Z. what is the total displacement of the car?

The car drives the legs a and b of a right triangle.
The resulting displacement is the hypotenuse c of the triangle.
The direction is N (60 ° - atan (50/120))E.

c = √(a² + b²) = √((120² + 50²)km²) = 130km

The direction is N (60° - atan(50/120))E = N37°23'E

!

(7-2/x)=sqrt(7/x)

\(7-\frac{2}{x}=\sqrt{\frac{7}{x}}\\ (7-\frac{2}{x})^2=\frac{7}{x}\\ 49-\frac{28}{x}+(\frac{2}{x})^2=\frac{7}{x}\\ 49+\frac{4}{x^2}=\frac{35}{x}\\ 49x^2+4=35x\\ 49x^2-35x+4=0\\ x=\frac{35\pm\sqrt{1225-16*49}}{98}\\ x=\frac{35\pm21}{98}\\ x=\frac{56}{98}\qquad or \qquad x=\frac{14}{98}\\ x=\frac{4}{7}\qquad or \qquad x=\frac{1}{7}\\ \)

but  

\(x>0\;\;and \;\;\\ 7-\frac{2}{x}>0\\ 7>\frac{2}{x}\\ 7x>2\\ x>2/7\\~\\ \therefore x=\frac{4}{7}\)

If you want to multiply fractions, you multiply the numerator by numerator and denominator by denominator .

So in this case, you multiply 1 by 1 and 2 by 2. And then you put the numbers back in place. It should look like this. 1/4

Solve for x:
7-2/x = sqrt(7) sqrt(1/x)

Multiply both sides by x:
7 x-2 = sqrt(7)/sqrt(1/x)

Multiply both sides by sqrt(1/x):
7/sqrt(1/x)-2 sqrt(1/x) = sqrt(7)

Square both sides:
(7/sqrt(1/x)-2 sqrt(1/x))^2 = 7

(7/sqrt(1/x)-2 sqrt(1/x))^2 = 49 x-28+4/x:
49 x-28+4/x = 7

Bring 49 x-28+4/x together using the common denominator x:
(49 x^2-28 x+4)/x = 7

Multiply both sides by x:
49 x^2-28 x+4 = 7 x

Subtract 7 x from both sides:
49 x^2-35 x+4 = 0

The left hand side factors into a product with two terms:
(7 x-4) (7 x-1) = 0

Split into two equations:
7 x-4 = 0 or 7 x-1 = 0

Add 4 to both sides:
7 x = 4 or 7 x-1 = 0

Divide both sides by 7:
x = 4/7 or 7 x-1 = 0

Add 1 to both sides:
x = 4/7 or 7 x = 1

Divide both sides by 7:
x = 4/7 or x = 1/7

7-2/x ⇒ 7-2/(1/7) = -7
sqrt(7) sqrt(1/x) ⇒ sqrt(7) sqrt(1/(1/7)) = 7:
So this solution is incorrect

7-2/x ⇒ 7-2/(4/7) = 7/2
sqrt(7) sqrt(1/x) ⇒ sqrt(7) sqrt(1/(4/7)) = 7/2:
So this solution is correct

The solution is:
Answer: | x = 4/7

Thanks, geno3141.

I actually managed to figure this out (in the exact same way) while I was going out to do errands, but thanks for verifying the solution.

I am curious about the case where x  <  -1  and x  >  1. Obviously this is a contradiction, but what does it mean? How can this contradiction be resolved? Or does it not need to be resolved, and it suffices to only have one half of the method provide a relevant answer?

Welcome!

The first step is to remember the order of operations, PEMDAS, or Parentheseis, Exponents, Multiplication and Division, and finally Addition and Subtraction.

If this is 21 * (4 * 5) - 23, then here's the steps:

21 * (4 * 5) - 23

Multiply the numbers within the parenthesis:

21 * (4 * 5) - 23

21 * (20) - 23

Multiply:

21 * (20) - 23

420 - 23

And hopefully the last step is easy enough, subtract:

420 - 23 = your answer

So, we have a square with an area of 392, and we want to find the side length.

The first thing we have to ask ourselves is the question, "what is a square?"

Thinking about it, the definition is "a shape that has four sides that are all the same size, and has right angles in each corner"

With that out of the way, we can realize how to find the area of a square from its side lengths: multiplication, in this case an operation aptly called "squaring".

By multiplying two perpindicular sides (represented by s) of the square, we arrive at the area (represented by a):

\(s \times s = a\), also represented as \({s}^{2} = a\)

Since we know what the area is, 392, we can substitute that for a:

\({s}^{2} = a\)

\({s}^{2} = 392\)

Then to cancle the "squared" operation, a square root is used:

\(\sqrt{s^2} = \sqrt{392}\)

\(s = \sqrt{392}\)

And there is the answer. The length of a side is \(\sqrt{392}\), which is approximately 19.7989898732233307

ElectricPavlov is correct.

One way to analyze the problem:  

     Each worker worked 10 eight-hour days for a total of 80 hours (10 x 8).

     Since there were four workers, they worked a total of 320 hours.

If you want to build the stage in 5 eight-hour days, each worker needs to work 40 hours (5 x 8).

The total number of hours needed is still 320 hours; since each worker works 40 hours, calculate  320 / 40 to get 8.

You will need 8 workers. 

26 x 25 x 24 = 15,600 combos

To get it done in HALF the time will take TWICE as many workers.      8 workers.

Did you mean   2  1/2 inches per side?

2 1/2  x  2  1/2  =  2.5 x 2.5 = 6.25 sq inches

Combination:     _nC_r  =  n! / [ (n - r)! · r! ]

If you assuming that only one number is chosen at a time, if there are  n  numbers, your answer will be  n!. 

Permutation: A set of objects in which position (or order) is
important=nPr=26P3 =15,600

Combination: A set of objects in which position (or order) is NOT
important.=nCr=26C3 =2,600

There are several ways to work this problem; this is one way:

                                                              x² < 1

Subtract 1 from both sides:              x² - 1 < 0

Factor:                                  (x + 1)(x - 1)  < 0

Analysis:  If two numbers are multiplied together and their answer is negative (less than 0), then either the first number is

                positive and the second number is negative or the first number is negative and the second number is positive.

                For this problem, the two numbers are  x + 1  and  x - 1.

So: first number positive, second number negative     or    first number negative, second number positive

                      (x + 1) >  0   and  (x - 1) <  0                            (x + 1) < 0   and  (x - 1)  > 0

                       x + 1  >  0    and   x - 1  <  0                              x + 1  < 0   and    x - 1  >  0

                             x  >  -1   and        x  < 1                                     x  <  -1  and        x  >  1

                                    -1  <  x  <  1                                                   <this is impossible>

Thus:     -1  <  x  <  1

It will still be imaginary.

Simple example:

1  x  i   =   i          

there needs to be another ' i ' to make the i-term go away.....

i  x  i  = i^2 =  - 1   

The formula for permutations is:  _nP_r  =  n! / (n - r)!

So:                  ₃₀P₂  =  30! / (30 - 2)!  =  30! / 28!

                                                           =  [ 30·29·28·27·26·...·4·3·2·1 ] / [ 28·27·26·...·4·3·2·1 ]

(which, after cancelling, becomes)     =  30·29  =  870

5280 ft  /  50 ft/3turns  =  5280 x 3/50 turns = 316.8 turns

To multiply fractions, you multiply the numerator and denominator of each.

So \({1 \over 2} \times {1 \over 2} = {1 \times 1 \over 2 \times 2} = {1 \over 4}\)

(1/6)²⁸  =  1 / 6, 140, 943, 214, 000, 000, 000, 000  (approx)

            =  1 in 6.14 x 10²¹  (approx)

However a septillion is 1 in 10²⁴, so (1/6)²⁸ misses a septillion by (about) a factor of 1,000.

(2w)² · (5w)²  =  (2xw) · (2w) · (5w) · (5w)  =  (2 · 2 · 5 · 5) · (w · w · w · w)  =  100w⁴

-6[6^2(55-6^5)/3]       =

-6 [ 36 * (55 - 6^5) / 3 ]  =

-6 [ 36 * ( 55  - 7776) / 3 ]  =

-6 [  36 * ( -7721) / 3 ]  =

-6 [ -277,956 / 3]   =

-6 [ - 92652]   =

555,912